I am trying to prove, or understand the proof of, the following part of the sylow theorems (according to my notes anyway):

Let G be a finite group and let p be a prime which divides the order of G. Then

Every p-subgroup of G is contained in a Sylow p-subgroup of G.

Proof: Let P be a Sylow p-subgroup. Let Q be any p-subgroup. Let be the set of all conjugates of P. Then G acts on by conjugation. By the orbit stabiliser theorem, the number of conjugates of P is r = | | = [G: ]. Note that p does not divide r.

but I don't understand why p does not divide r! could anyone help me with out with this? thanks very much

April 16th 2010, 06:51 PM

aliceinwonderland

Quote:

Originally Posted by slevvio

I am trying to prove, or understand the proof of, the following part of the sylow theorems (according to my notes anyway):

Let G be a finite group and let p be a prime which divides the order of G. Then

Every p-subgroup of G is contained in a Sylow p-subgroup of G.

Proof: Let P be a Sylow p-subgroup. Let Q be any p-subgroup. Let be the set of all conjugates of P. Then G (P?) acts on by conjugation. By the orbit stabiliser theorem, the number of conjugates of P is r = | | = [G: ]. Note that p does not divide r.

but I don't understand why p does not divide r! could anyone help me with out with this? thanks very much

Lemma . If a group H of order ( p prime) acts on a finite set and if then (Hungerford, p93).

Let be the set of all conjugates of P and let P acts on by conjugation. An element T (sylow P-subgroup) in satisfies in the above lemma iff for all . We see that . We also know that T is normal in . By the second Sylow theorem, P and T should be conjuate in . This forces T= P and . Since and p does not divide , we conclude that .

April 17th 2010, 04:18 AM

slevvio

hello, thanks for your response, however I dont think i was very clear. This proof is meant to be the first part in proving the conjugacy theorem.

So far I have proved :

1) if the order of G is , where p does not divide s, then there exists a Sylow-p subgroup of G

2) there exists a subgroup of G of any p-power order, as long as that power of p divides .

Now I am trying to show that 3) Any p-subgroup of G is contained in a Sylow p-subgroup of G (I think this is sayign that that subgroup is maximal)

So of course I haven't proved the conjugacy theorem yet. I will show you the exact way the proof is written out in the notes, the way they say that p does not divide r seems as if it is almost offhand

(iii) Let
P be a Sylow p-subgroup. Let Q be any p-subgroup. Let be the set of all
conjugates of P. Then G acts on by conjugation. By the orbit–stabilizer theorem, the number
r of conjugates of P is . Note that . Then Q acts on by conjugation. Each
Q-orbit has size a power of P. There must be an orbit whose order is not divisible by p. However the only power of p which is not divisible by p is = 1. So there is an orbit of size 1. Let element in an orbit of size 1. Then Q normalizes . Therefore is a p-group. Hence and so .
(iv) Assume now that
Q is a Sylow p-subgroup. Then it follows that . Thus every Sylow p-subgroup is conjugate to P.

I don't realy understand any of it yet but I am stuck on the p does not divide r part

April 17th 2010, 02:11 PM

aliceinwonderland

Quote:

Originally Posted by slevvio

hello, thanks for your response, however I dont think i was very clear. This proof is meant to be the first part in proving the conjugacy theorem.

So far I have proved :

1) if the order of G is , where p does not divide s, then there exists a Sylow-p subgroup of G

2) there exists a subgroup of G of any p-power order, as long as that power of p divides .

Now I am trying to show that 3) Any p-subgroup of G is contained in a Sylow p-subgroup of G (I think this is sayign that that subgroup is maximal)

So of course I haven't proved the conjugacy theorem yet. I will show you the exact way the proof is written out in the notes, the way they say that p does not divide r seems as if it is almost offhand

(iii) Let
P be a Sylow p-subgroup. Let Q be any p-subgroup. Let be the set of all
conjugates of P. Then G acts on by conjugation. By the orbit–stabilizer theorem, the number
r of conjugates of P is . Note that . Then Q acts on by conjugation. Each
Q-orbit has size a power of P. There must be an orbit whose order is not divisible by p. However the only power of p which is not divisible by p is = 1. So there is an orbit of size 1. Let element in an orbit of size 1. Then Q normalizes . Therefore is a p-group. Hence and so .
(iv) Assume now that
Q is a Sylow p-subgroup. Then it follows that . Thus every Sylow p-subgroup is conjugate to P.

I don't realy understand any of it yet but I am stuck on the p does not divide r part

In case G acts on by conjugation, there is only one orbit in under G such that , where and , p does not divide s. Here, r cannot be equal to p and , so .